Freezing Point Depression Given: Mass (g) of p-Dichlorobenzene used: 9.67 Initia

Question

Freezing Point Depression

Given:

Mass (g) of p-Dichlorobenzene used: 9.67

Initial Mass (g) of opened sample tube: 13.9023

Mass (g) (Sample tube - Sample 1): 13.2575

Mass (g) of sample tube with remaining sample: 13.2575

Mass (g) of sample tube - both samples: 12.7347

Freezing point (°C) of pure p-Dichorobenzene: 52.9

Freezing point (°C) of Solution 1: 50.2

Freezing point (°C) of Solution 2: 48.2

Question:

1. What is Mass (g) of UNKNOWN added to Solution 1?
2. What is Total mass (g) of UNKNOWN in Solution 2?
3. What is Freezing Point Depression (°C) of Solution 1?
4. What is Freezing Point Depression (°C) of Solution 2?
5. What is Unknown molecular weight (g/mol) calculated from pure PDB and Solution 1?
6. What is Unknown molecular weight (g/mol) calculated from pure PDB and Solution 2?
7. What is Unknown average molecular weight (g/mol)?

Edit: The name of the solvent that is being used is p-Dichlorobenzene, which is stated in the givens. Its' Kf value is not given to me. 1st question is asking for the amount in terms of grams that is added to solution 1. 2nd question is asking the overall mass of unknown in solution 2.

Explanation / Answer

Mass of sample tube = 13.9023 g

mass of p-dichloro benzene = 9.67 g

mass of sample-1 = 13.9023 - 13.2575 = 0.6448 g

mass of sample-1 and 2 = 13.9023 - 12.7347 = 1.1676 g

mass of sample-2 = 1.1676 - 0.6448 = 0.5228g

mass of solution-1 = mass of solvent + mass of sample-1

mass of solution-1 = 9.67 + 0.6448 = 10.3148 g

mass of solution-2 = mass of solvent + mass of sample-2

mass of solution-2 = 9.67 + 0.5228 = 10.1928g

Freezing point of depression of solution-1 = 52.9 - 50.2 = 2.7 degree celcius

Freezing point of depression of solution-2 = 52.9 - 48.2 = 4.7 degree celcius.

we have the formula,

Depression in freezing point = Kf * molalaity of the solution

for sample-1,

2.7 = 7.1 * (0.6448/M)*(1000/9.67) where M = molar mass of sample-1

M = molar mass = 175.35 g/mol

for sample-2,

4.7 = 7.1 * (0.5228/M)*(1000/9.67) where M = molar mass of sample-2

M = molar mass = 81.67 g/mol

average molecular weight = (175.35 + 81.67)/2 = 128.51 g/mol

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